Solve for $y$, $ -\dfrac{3}{25y + 10} = -\dfrac{10}{10y + 4} + \dfrac{4y - 1}{5y + 2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25y + 10$ $10y + 4$ and $5y + 2$ The common denominator is $50y + 20$ To get $50y + 20$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{3}{25y + 10} \times \dfrac{2}{2} = -\dfrac{6}{50y + 20} $ To get $50y + 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{10}{10y + 4} \times \dfrac{5}{5} = -\dfrac{50}{50y + 20} $ To get $50y + 20$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ \dfrac{4y - 1}{5y + 2} \times \dfrac{10}{10} = \dfrac{40y - 10}{50y + 20} $ This give us: $ -\dfrac{6}{50y + 20} = -\dfrac{50}{50y + 20} + \dfrac{40y - 10}{50y + 20} $ If we multiply both sides of the equation by $50y + 20$ , we get: $ -6 = -50 + 40y - 10$ $ -6 = 40y - 60$ $ 54 = 40y $ $ y = \dfrac{27}{20}$